Prove that BD bisects angle ABC Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)A geometry problem that involves congruence of triangles.Prove that: $S_XYZgeq frac14S_ABC$Prove that angle ACB > angle ABD.Let $D, E, F$ be the feet of the altitudes from $A, B, C$ in $triangle ABC$. Prove that the perpendicular bisector of $EF$ also bisects $BC$.In the following figure, prove that $AC$ bisects $GH$.In triangle $ABC$ find angle $angle BAC$ given that…Show that the altitude bisects the corresponding angleAngle bisector contains the Nine Point CentreProve sum of angles in problem involving bisectors in a given triangleHow to solve for $angle BDC$ given the information of other angles in the picture

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Prove that BD bisects angle ABC



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)A geometry problem that involves congruence of triangles.Prove that: $S_XYZgeq frac14S_ABC$Prove that angle ACB > angle ABD.Let $D, E, F$ be the feet of the altitudes from $A, B, C$ in $triangle ABC$. Prove that the perpendicular bisector of $EF$ also bisects $BC$.In the following figure, prove that $AC$ bisects $GH$.In triangle $ABC$ find angle $angle BAC$ given that…Show that the altitude bisects the corresponding angleAngle bisector contains the Nine Point CentreProve sum of angles in problem involving bisectors in a given triangleHow to solve for $angle BDC$ given the information of other angles in the picture










6












$begingroup$



Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac12BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here










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  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    1 hour ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    1 hour ago















6












$begingroup$



Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac12BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here










share|cite|improve this question









New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    1 hour ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    1 hour ago













6












6








6





$begingroup$



Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac12BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here










share|cite|improve this question









New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





Given that $triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=frac12BD$, prove that BD bisects angle $angle ABC$.




I have tried proving triangle $triangle AEB$ and triangle $triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here







geometry triangles






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New contributor




Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited 1 hour ago







Pushpa Kumari













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asked 2 hours ago









Pushpa KumariPushpa Kumari

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New contributor





Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Pushpa Kumari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    1 hour ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    1 hour ago
















  • $begingroup$
    Start by drawing a diagram and showing all the given information.
    $endgroup$
    – 1123581321
    2 hours ago










  • $begingroup$
    @PushpaKumari just provide a link to your image, someone will be willing to edit it.
    $endgroup$
    – Quang Hoang
    1 hour ago










  • $begingroup$
    To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
    $endgroup$
    – 1123581321
    1 hour ago















$begingroup$
Start by drawing a diagram and showing all the given information.
$endgroup$
– 1123581321
2 hours ago




$begingroup$
Start by drawing a diagram and showing all the given information.
$endgroup$
– 1123581321
2 hours ago












$begingroup$
@PushpaKumari just provide a link to your image, someone will be willing to edit it.
$endgroup$
– Quang Hoang
1 hour ago




$begingroup$
@PushpaKumari just provide a link to your image, someone will be willing to edit it.
$endgroup$
– Quang Hoang
1 hour ago












$begingroup$
To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
$endgroup$
– 1123581321
1 hour ago




$begingroup$
To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite.
$endgroup$
– 1123581321
1 hour ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Refer to the figure:



$hspace2cm$enter image description here



From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
$$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
From the right $Delta BCD$:
$$z^2+y^2=(2x)^2 (2)$$
Now substitute $(1)$ to $(2)$:
$$z^2+y^2=2(y^2-zy) Rightarrow \
(y-z)^2=2z^2 Rightarrow \
y-z=zsqrt2 Rightarrow \
fracy-zz=fracysqrt2y,$$

which is consistent with the angle bisector theorem.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    enter image description here



    Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



    First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



    Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
    Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      A simple geometric solution:



      Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac 12DB=frac 12AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Refer to the figure:



        $hspace2cm$enter image description here



        From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
        $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
        From the right $Delta BCD$:
        $$z^2+y^2=(2x)^2 (2)$$
        Now substitute $(1)$ to $(2)$:
        $$z^2+y^2=2(y^2-zy) Rightarrow \
        (y-z)^2=2z^2 Rightarrow \
        y-z=zsqrt2 Rightarrow \
        fracy-zz=fracysqrt2y,$$

        which is consistent with the angle bisector theorem.






        share|cite|improve this answer









        $endgroup$

















          2












          $begingroup$

          Refer to the figure:



          $hspace2cm$enter image description here



          From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
          $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
          From the right $Delta BCD$:
          $$z^2+y^2=(2x)^2 (2)$$
          Now substitute $(1)$ to $(2)$:
          $$z^2+y^2=2(y^2-zy) Rightarrow \
          (y-z)^2=2z^2 Rightarrow \
          y-z=zsqrt2 Rightarrow \
          fracy-zz=fracysqrt2y,$$

          which is consistent with the angle bisector theorem.






          share|cite|improve this answer









          $endgroup$















            2












            2








            2





            $begingroup$

            Refer to the figure:



            $hspace2cm$enter image description here



            From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
            $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
            From the right $Delta BCD$:
            $$z^2+y^2=(2x)^2 (2)$$
            Now substitute $(1)$ to $(2)$:
            $$z^2+y^2=2(y^2-zy) Rightarrow \
            (y-z)^2=2z^2 Rightarrow \
            y-z=zsqrt2 Rightarrow \
            fracy-zz=fracysqrt2y,$$

            which is consistent with the angle bisector theorem.






            share|cite|improve this answer









            $endgroup$



            Refer to the figure:



            $hspace2cm$enter image description here



            From similarity of triangles $Delta ADE$ and $Delta BCD$ (corresponding angles are equal):
            $$fracxy=fracy-z2x Rightarrow 2x^2=y^2-zy (1)$$
            From the right $Delta BCD$:
            $$z^2+y^2=(2x)^2 (2)$$
            Now substitute $(1)$ to $(2)$:
            $$z^2+y^2=2(y^2-zy) Rightarrow \
            (y-z)^2=2z^2 Rightarrow \
            y-z=zsqrt2 Rightarrow \
            fracy-zz=fracysqrt2y,$$

            which is consistent with the angle bisector theorem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            farruhotafarruhota

            22.3k2942




            22.3k2942





















                2












                $begingroup$

                enter image description here



                Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



                First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



                Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
                Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






                share|cite|improve this answer









                $endgroup$

















                  2












                  $begingroup$

                  enter image description here



                  Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



                  First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



                  Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
                  Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






                  share|cite|improve this answer









                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    enter image description here



                    Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



                    First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



                    Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
                    Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.






                    share|cite|improve this answer









                    $endgroup$



                    enter image description here



                    Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.



                    First notice that $angle GMD = angle DAE = 90^circ -angle ADE$.



                    Now $triangle MGD$ and $triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$
                    Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $angle ABC$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Quang HoangQuang Hoang

                    13.3k1233




                    13.3k1233





















                        0












                        $begingroup$

                        A simple geometric solution:



                        Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac 12DB=frac 12AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          A simple geometric solution:



                          Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac 12DB=frac 12AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            A simple geometric solution:



                            Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac 12DB=frac 12AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.






                            share|cite|improve this answer









                            $endgroup$



                            A simple geometric solution:



                            Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=frac 12DB=frac 12AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 10 mins ago









                            siroussirous

                            1,7481514




                            1,7481514




















                                Pushpa Kumari is a new contributor. Be nice, and check out our Code of Conduct.









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